Question

In triangle $ABC,$ point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2.$ Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE.$ Given that the area of $\triangle ABC$ is $360,$ what is the area of $\triangle EBF?$

$24$

$30$

$32$

$36$

$40$

Video solution:

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Written solution:

Since $AD:DC=1:2,$ triangles $ABD$ and $DBC$ have areas in the ratio $1:2.$ Thus $[ABD]=120$ and $[DBC]=240.$

Because $E$ is the midpoint of $BD,$ triangles $ABE$ and $AED$ each have area $60.$ Let $x=[EBF].$ Then $[DEF]=x$ as well, since $BE=ED$ and both triangles have their third vertex on line $AF.$

Segment $DF$ splits $\triangle DBC,$ so $[DFC]=240-2x.$ Also, $\triangle ADF$ and $\triangle DFC$ have bases $AD$ and $DC$ on the same line, so their areas are in the ratio $1:2.$

Therefore $\dfrac{60+x}{240-2x}=\dfrac{1}{2}.$ Solving gives $120+2x=240-2x,$ so $x=30.$

Thus, the correct answer is B.

2019 AMC 8 Problem 24

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Written solution:

Problem 24 in Other Years